Orthogonal Curvilinear Coordinates - Research

CBE 6333, R. Levicky 1

Orthogonal Curvilinear Coordinates

Introduction. Rectangular Cartesian coordinates are convenient when solving problems in which the
geometry of a problem is well described by the coordinates x1, x2, and x3. For example, in Fig. 1, the
geometry of the flow is easily described by saying that it occurs in the x1 direction, with the velocity v
= v1 δ1 being zero at the boundaries x2 = 0 and x2 = d (i.e. no-slip boundary condition). But what about
the flow in Fig. 2? Here, the flow could be described by stating that it occurs in the θ direction, with vθ
equal to zero at r = r1 and to ω r2 at r = r2. Such a description uses cylindrical, not Cartesian,
coordinates. It is not necessary to use cylindrical coordinates, but their use does simplify the
description (and mathematical solution) of a problem like the one in Fig. 2. If instead the flow in Figure
2 is described in Cartesian coordinates, the description becomes more convoluted and may go as
follows: the components v1 and v2 of the velocity v1 δ1 + v2 δ2 add so as to make a fluid element travel
in a circle, with both components equal to zero at (x12 + x22)1/2 = r1 and obeying the condition v12 + v22
= ω2 r22 at (x12 + x22)1/2 = r2. This description is correct, but solving a problem possessing a cylindrical
geometry using Cartesian coordinates will be much more cumbersome.

x2

x2 v1(x2) d
x1

Figure 1. Flow between two flat plates. r2 r1

x1

Figure 2. Circular flow in an annulus. The outer vθ (r)
wall of the annulus is rotating with an angular
velocity ω. ω

For geometries as in Fig. 2, proper use of "orthogonal curvilinear coordinates" can simplify the solution
to the problem. What are orthogonal curvilinear coordinates? The most familiar examples (there are
many others) are cylindrical and spherical coordinates as illustrated in Figures 3 and 4. The cylindrical
and spherical coordinate systems are termed "curvilinear" because some of the coordinates change
along curves. For instance, in cylindrical coordinates, θ changes along a curve that can be thought of as
forming a circle about the origin. The Cartesian coordinate system is not curvilinear since all of the
coordinates change along straight lines. The curve along which one coordinate changes while the other
coordinates remain fixed is a coordinate curve for that coordinate. Coordinate curves for θ in the
cylindrical coordinate system describe circles around the origin, while those for r are lines that radiate
outward from the origin. Two coordinate curves for θ and one for r are shown in Figure 3. Coordinate
curves for θ in the spherical geometry (Figure 4) describe semicircles (0 < θ < 180o), while those for φ
describe circles about the origin. In addition to being curvilinear, the cylindrical and spherical
coordinate systems are also "orthogonal" because coordinate curves corresponding to different
coordinates are perpendicular to one another. For example, in cylindrical coordinates, the coordinate

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curves for r are perpendicular to those for θ and x3. The coordinate curves for θ are perpendicular to
those for r and x3, etc. Clearly, the Cartesian coordinate system is also orthogonal.

x3 x3

x2 r
r x2

x1 x1

Figure 3. Cylindrical coordinates. The Figure 4. Spherical coordinates. The
coordinates are r, θ and x3. coordinates are r, θ and φ.

Unit basis vectors are defined at each point in the coordinate system as sketched in Figures 5 and 6. In
general, the direction of a basis vector changes with position. In cylindrical coordinates, both δr and δθ
change direction with position as illustrated in Fig 5, while the direction of δx3 is independent of
position. In the spherical coordinate system, all three basis vectors δr, δθ and δφ change direction with
position. The magnitudes of the basis vectors do not change, since by definition the length of a basis
vector is always unity.

x3

x3

x2 x2

x1 x1

Figure 5. Basis vectors for the cylindrical coordinate Figure 6. Basis vectors for the spherical
system. coordinate system.

An important fact to recall: the length s of an arc on a circle is given by the product of the circle's
radius r and the angle ζ (ζ is in radians) through which the arc sweeps; s = rζ , see Figure 7. This
relation, especially in its differential form ds = r dζ, will be useful in the following discussion.

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Figure 7.

Transformation of Coordinates.
The coordinates of a point in space, expressed in terms of two different coordinate systems, can be
related by transformation of coordinates equations. Let's say the first coordinate system employs q1,
q2, and q3 as coordinates, and the second employs x1, x2, and x3. Then the transformation of coordinates
equations have the general form

x1 = x1(q1, q2, q3) x2 = x2(q1, q2, q3) x3 = x3(q1, q2, q3) (1)

and

q1 = q1(x1, x2, x3) q2 = q2(x1, x2, x3) q3 = q3(x1, x2, x3) (2)

For example, if x1, x2, and x3 are the Cartesian coordinates and q1, q2, and q3 are the cylindrical
coordinates r, θ, and z, then

x1 = r cosθ x2 = r sinθ x3 = x3 (3)

and

r = (x12 + x22)1/2 θ = tan-1(x2/ x1) x3 = x3 (4)

Equations 3 and 4 are the transformation of coordinates equations between the Cartesian and the
cylindrical coordinate systems. For spherical coordinates, the transformation equations are

x1 = r sinθ cosφ x2 = r sinθ sinφ x3 = r cosθ (5)

and

r = (x12 + x22 + x32)1/2 θ = cos-1(x3 / (x12 + x22 + x32)1/2) φ = tan-1(x2/ x1) (6)

Basis Vectors. The basis vectors possess unit magnitude and point along the direction of coordinate
curves as illustrated in Figures 5 and 6. Because the Cartesian basis vectors are constant in direction as
well as magnitude, expressing curvilinear basis vectors in terms of the Cartesian ones can simplify
mathematical derivations in some instances. Therefore, it is desired to derive expressions for the

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curvilinear basis vectors in terms of the Cartesian basis. To begin, we first define the position p of a
point in space in terms of a Cartesian coordinate system:

p = x1(q1, q2, q3) δ1 + x2(q1, q2, q3) δ2 + x3(q1, q2, q3) δ3 (6b)

Expressions for the Cartesian coordinates xi in expression 6b in terms of the cylindrical coordinates are
given by equations 3; for the spherical coordinate system, by equations 5. A unit basis vector δi in the
direction of the coordinate qi can be expressed as

∂p ∂p (7)
δ i = ∂qi ∂qi

∂p  ∂p ∂p  1/ 2 ∂p ∂p
∂qi  ∂qi ∂qi  ∂qi ∂qi
where = ⋅ is the magnitude of the vector . Note that the vector is

tangent to (i.e. points along) the coordinate curve for qi (Figure 8). The division by the magnitude

∂p ∂p ∂p
scales the vector to unit magnitude; thus, is often referred to as the "scale factor" hi.
∂qi ∂qi ∂qi

The result of equation (7) is a unit basis vector whose direction points along the coordinate curve for qi

∂p
and whose length is normalized to unity by dividing by ; in other words, this is the unit basis

∂qi

vector corresponding to the coordinate qi.

Figure 8

Equation (7) works trivially for the Cartesian coordinate system. In the Cartesian system, p = x1 δ1 + x2
δ2 + x3 δ3 so that, for example, δ1 = (∂p/∂x1) / |(∂p/∂x1)| = δ1 / (δ1•δ1)1/2 = δ1 / 1 = δ1. Here, the scale
factor h1 = (δ1•δ1)1/2 = 1. What about the cylindrical coordinate system? In the cylindrical system, using

equations (3)

p = r cosθ δ1 + r sinθ δ2 + x3 δ3 (8)

Then,

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δr = (∂p/∂r) / |(∂p/∂r)| = (cosθ δ1 + sinθ δ2) / [(cosθ δ1 + sinθ δ2)• (cosθ δ1 + sinθ δ2)]1/2
= (cosθ δ1 + sinθ δ2) / [cos2θ + sin2θ]1/2

δr = cosθ δ1 + sinθ δ2 (9)

Note that the scale factor hr = [(cosθ δx + sinθ δy)• (cosθ δx + sinθ δy)]1/2 = [cos2θ + sin2θ]1/2 = 1.
Similarly,

δθ = (∂p/∂θ) / |(∂p/∂θ)| (10)
= (- r sinθ δ1 + r cosθ δ2) / [(- r sinθ δ1 + r cosθ δ2) • (- r sinθ δ1 + r cosθ δ2)]1/2
= (- r sinθ δ1 + r cosθ δ2) / [r2 sin2θ + r2 cos2θ]1/2
= (- r sinθ δ1 + r cosθ δ2) / [r2 (sin2θ + cos2θ)]1/2
= (- r sinθ δ1 + r cosθ δ2) / r

δθ = - sinθ δ1 + cosθ δ2

Note that the scale factor hθ = |(∂p/∂θ)| = r. To summarize, for the cylindrical coordinate system

δr = cosθ δ1 + sinθ δ2 δθ = - sinθ δ1 + cosθ δ2 δx3 = δ3 (11)

hr = 1 hθ = r hx3 = 1 (12)

By similar procedures it can be shown that for the spherical coordinate system,

δr = sinθ cos φ δ1 + sinθ sin φ δ2 + cosθ δ3 δθ = cosθ cos φ δ1 + cosθ sin φ δ2 - sinθ δ3

δφ = - sin φ δ1 + cos φ δ2 (13)

hr = 1 hθ = r hφ = r sinθ (14)

From expressions 11 and 13 it should be clear that the direction of δr and δθ in the cylindrical system,
and the direction of all three basis vectors in the spherical coordinate system, changes with position.

Expressions 11 and 13 are the desired equations that express the curvilinear basis vectors in terms of

their Cartesian counterparts. How do we apply these equations? For instance, an angular velocity v = vθ
δθ expressed in the cylindrical coordinate system can be converted to v = - vθ sinθ δ1 + vθ cosθ δ2 by
using equation 11 for δθ. If the expression θ = tan-1(x2/x1) is also substituted (see equations 4), the
velocity will then be purely expressed in terms of Cartesian coordinate variables and basis vectors.

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Length, Area and Volume Elements. For a set of orthogonal curvilinear coordinates q1, q2 and q3,
with corresponding scaling factors h1, h2, and h3, a differential displacement in position dp is given by
(note the use of the summation convention)

dp = h1 dq1 δ1 + h2 dq2 δ2 + h3 dq3 δ3 = hi dqi δi (15)

For the Cartesian coordinate system, all of the scale factors equal one, and equation (15) becomes

dp = dx1 δ1 + d x2 δ2 + dx3 δ3 (16)

as seen previously. For the cylindrical coordinate system, using equations (12) for the scale factors,

dp = dr δr + r dθ δθ + dx3 δx3 (17)

Why are the scale factors hi needed in equation 15 for dp? Earlier, it was mentioned (Figure 7) that the
length of an arc on the circumference of a circle is equal to the product of the central angle that spans
the arc times the radius of the circle on whose circumference the arc lies. Therefore, the distance that is
traversed when θ changes by an infinitesimal amount dθ is equal to rdθ (Figure 9). Since dp expresses
a change in distance, rdθ must be used in equation 17 rather than just simply dθ (dθ is a change in
angular position, and is not a distance). The distance corresponding to differential changes in the
various coordinates is:

x3

x1 Figure 9. An object initially at position p is displaced by rdθ δθ to a final
position p + rdθ δθ.
dθ
x2

rdθ

rdθ

Table I. Coordinate Distance corresponding to an
Coordinate System infinitesimal change in coordinate
x1
Cartesian x2 dx1
x3 dx2
Cylindrical dx3
r
θ dr
x3 rdθ
dx3

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Spherical
r dr
θ rdθ
φ r sinθ dφ

The volume dV of an infinitesimal volume element is obtained as usual; i.e. it is given by the product
of the lengths of the three sides defining the length, depth and width of the element. Each of the three
sides of the volume element is taken to lie along one of the coordinate directions. Note that the sides
are assured to be mutually orthogonal since only orthogonal coordinate systems are being considered.
In general

dV = h1 h2 h3 dq1 dq2 dq3 (18)

For the Cartesian, cylindrical and spherical systems, equation 18 evaluates to

Table II.

Coordinate System dV

Cartesian dx1 dx2 dx3 (19a)
Cylindrical dr rdθ dx3 = r dr dθ dx3 (19b)
Spherical dr rdθ rsinθ dφ = r2sinθ dr dθ dφ (19c)

By similar reasoning, the area dA of an infinitesimal area element is given by the product of the lengths
of its sides. For instance, a differential area element in the Cartesian x1-x2 plane is dA = dx1 dx2. A
differential area element in the θ-x3 cylindrical surface is r dθ dx3 (Figure 10). A differential area
element in the θ-φ spherical surface is r2sinθ dθ dφ, etc.

x3
x3

x2 dθ dx3rdθ x2

x1 dx1 dA = dx1dx2 dA = rdθdx3
dx2 x1 Figure 10.

Derivatives of Basis Vectors with Respect to Coordinate Variables.
As pointed out earlier (Figures 5 and 6), in general the direction of a basis vector can change from
point to point. If a change in coordinate position can cause a change in the direction of a basis vector,
that means that the basis vector must have a nonzero derivative with respect to at least some of the

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coordinate variables qi. For example, in the cylindrical coordinate system the basis vectors can be
written (equations 11)

δr = cosθ δ1 + sinθ δ2 δθ = - sinθ δ1 + cosθ δ2 δ3 = δ3 (11)

The derivative of δr with respect to θ then becomes

∂δr / ∂θ = - sinθ δ1 + cosθ δ2 = δθ.

Therefore, the derivative ∂δr / ∂θ is nonzero. By similar calculations, it can be shown (you should
verify these equations by directly considering expressions 11 above):

∂δr / ∂r = 0 ∂δr / ∂θ = δθ ∂δr / ∂x3 = 0

∂δθ / ∂r = 0 ∂δθ / ∂θ = - δr ∂δθ / ∂x3 = 0 (20)
∂δx3 / ∂r = 0 ∂δx3 / ∂θ = 0 ∂δx3 / ∂x3 = 0

For the spherical coordinate system, the basis vectors are given by equations 13

δr = sinθ cos φ δ1 + sinθ sin φ δ2 + cosθ δ3 δθ = cosθ cos φ δ1 + cosθ sin φ δ2 - sinθ δ3
δφ = - sin φ δ1 + cos φ δ2 (13)

The derivatives of the basis vectors for the spherical coordinate system become

∂δr / ∂r = 0 ∂δr / ∂θ = δθ ∂δr / ∂φ = sinθ δφ

∂δθ / ∂r = 0 ∂δθ / ∂θ = - δr ∂δθ / ∂φ = cosθ δφ (21)
∂δφ / ∂r = 0 ∂δφ / ∂θ = 0 ∂δφ / ∂φ = - sinθ δr - cosθ δθ

Each of these expressions can be derived from equations 13.

Gradient, Divergence and Curl.
The gradient operator ∇ takes the derivative of a quantity with respect to a change in position. In
orthogonal curvilinear coordinates q1, q2 and q3, the ∇ operator is (note that the scaling factors hi in the
denominator ensure that a derivative is being taken with respect to distance)

1∂ 1∂ 1∂
∇ = δ1 h1 ∂q1 + δ 2 h2 ∂q2 + δ 3 h3 ∂q3 (22)
(23a)
Cartesian ∇ = δ1 1 ∂ + δ2 1 ∂ + δ3 1 ∂
1 ∂x1 1 ∂x2 1 ∂x3

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(23b)
Cylindrical ∇ = δ r 1 ∂ + δ θ 1 ∂ + δ x3 1 ∂ (23c)
Spherical 1 ∂r r ∂θ 1 ∂x3

1∂ 1∂ 1∂
∇ = δ r 1 ∂r + δθ r ∂θ + δ φ r sin θ ∂φ

The divergence of a vector A, written ∇•A, can be calculated using the definition of the gradient. For
the cylindrical coordinate system,

A = Ar(r, θ, x3)δr + Aθ( r, θ, x3)δθ + Ax3(r, θ, x3)δx3

and ∇•A =  ∂ + δθ 1 ∂ + δ x3 ∂  • [Ar(r, θ, x3)δr + Aθ( r, θ, x3)δθ + Ax3(r, θ, x3)δx3]
δ r ∂r r ∂θ ∂x3 

= δr • (∂/∂r) [Ar(r, θ, x3)δr + Aθ( r, θ, x3)δθ + Ax3(r, θ, x3)δx3] +

δθ • 1/r (∂/∂θ)[Ar(r, θ, x3)δr + Aθ( r, θ, x3)δθ + Ax3(r, θ, x3)δx3] +

δx3 • (∂/∂x3) [Ar(r, θ, x3)δr + Aθ( r, θ, x3)δθ + Ax3(r, θ, x3)δx3]

Using the product rule for differentiation

= δr • (δr ∂Ar/∂r + Ar ∂δr/∂r + δθ ∂Aθ/∂r + Aθ ∂δθ/∂r + δx3 ∂Ax3/∂r + Ax3 ∂δx3/∂r) +
δθ • 1/r (δr ∂Ar/∂θ + Ar ∂δr/∂θ + δθ ∂Aθ/∂θ + Aθ ∂δθ/∂θ + δx3 ∂Ax3/∂θ + Ax3 ∂δx3/∂θ) +
δx3 • (δr ∂Ar/∂x3 + Ar ∂δr/∂x3 + δθ ∂Aθ/∂x3 + Aθ ∂δθ/∂x3 + δx3 ∂Ax3/∂x3 + Ax3 ∂δx3/∂x3)

Substituting equations 20 for the derivatives of the basis vectors yields

= δr • (δr ∂Ar/∂r + δθ ∂Aθ/∂r + δx3 ∂Ax3/∂r) +
δθ • 1/r (δr ∂Ar/∂θ + Ar δθ + δθ ∂Aθ/∂θ - Aθ δr + δx3 ∂Ax3/∂θ ) +
δx3 • (δr ∂Ar/∂x3 + δθ ∂Aθ/∂x3 + δx3 ∂Ax3/∂x3)

Performing the dot products (remembering that δi • δj = δij since the basis vectors are mutually
orthogonal)

∇•A = ∂Ar/∂r + Ar/r + 1/r ∂Aθ/∂θ + ∂Ax3/∂x3 (24)

Expression 24 is the divergence of an arbitrary vector A in cylindrical coordinates. An analogous
approach could be used to derive ∇•A in spherical or Cartesian coordinates. The results are

∇•A = ∂A1/∂x1 + ∂A2/∂x2 + ∂A3/∂x3 (Cartesian) (25a)

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∇•A = ∂Ar/∂r + Ar/r + 1/r ∂Aθ/∂θ + ∂Ax3/∂x3 (cylindrical) (25b)
(25c)
∇•A = ∂Ar/∂r + 2Ar/r + 1/r ∂Aθ/∂θ + cosθ Aθ /(r sinθ) + 1/(r sinθ) ∂Aφ/∂φ (spherical)

It can be shown that, for an orthogonal curvilinear system, the divergence can be written as

1 ∂ ∂ ∂  (26)
∇•A = h1h2h3  ∂q1 (h2h3 A1) + ∂q2 (h1h3 A2 ) + ∂q3 (h1h2 A3)

where the hi are the scale factors for the chosen coordinate system (i.e. equations 12 and 14) and qi are
the coordinate variables. Equation 26 will work for any orthogonal curvilinear coordinate system, and
will reproduce equations 25a through 25c if the appropriate hi and qi are substituted into it.

Expressions for the curl ∇ × A of a vector A could be derived using a similar approach to that used in
deriving equation 24 for ∇ • A. Here, we simply write the final general formula for ∇ × A,

∇×A= 1   ∂ (h3A3) − ∂ (h2 A2) + h2δ 2  ∂ (h1A1) − ∂ (h3A3) + h3δ 3  ∂ (h2A2) − ∂ (h1A1)
h1h2h3 h1δ1 ∂q2 ∂q3 ∂q3 ∂q1 ∂q1 ∂q2

(27)

Equation 27 can be used to write out ∇ × A in the coordinate system of interest, provided that the
coordinate system is orthogonal curvilinear.

Concluding Remarks.
Some of the above expressions, even if written in simplified form, are rather cumbersome. Fortunately,
most of the equations needed for fluid dynamics have already been written down for the coordinate
systems of greatest interest, i.e. the Cartesian, cylindrical, and spherical systems. Therefore, it will not
be necessary to apply the above equations to express the Navier-Stokes equations in spherical
coordinates, for example. The desired expressions can be found in virtually any text on transport
phenomena.